Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 105;
int mar[N][N][2], g[N][N], n, m;
int rows[N][2], columns[N][2];
int ans[2];
void init() {
  memset(ans, 0, sizeof(ans));
  memset(mar, 0, sizeof(mar));
  memset(rows, 0, sizeof(rows));
  memset(columns, 0, sizeof(columns));
  memset(g, 0, sizeof(g));
}
void row_do(int x, int kind) {
  int min_g = g[x][0];
  int min_m = mar[x][0][kind];
  for (int i = 1; i < m; i++) {
    min_g = min(min_g, g[x][i]);
    min_m = min(min_m, mar[x][i][kind]);
  }
  rows[x][kind] = (min_g - min_m) > 0 ? (min_g - min_m) : 0;
  if (rows[x] == 0) return;
  ans[kind] += rows[x][kind];
  for (int i = 0; i < m; i++) mar[x][i][kind] += rows[x][kind];
}
void column_do(int y, int kind) {
  int min_g = g[0][y];
  int min_m = mar[0][y][kind];
  for (int i = 0; i < n; i++) {
    min_g = min(min_g, g[i][y]);
    min_m = min(min_m, mar[i][y][kind]);
  }
  columns[y][kind] = (min_g - min_m) > 0 ? (min_g - min_m) : 0;
  if (columns[y] == 0) return;
  ans[kind] += columns[y][kind];
  for (int i = 0; i < n; i++) mar[i][y][kind] += columns[y][kind];
}
void solve() {
  for (int i = 0; i < n; i++) {
    row_do(i, 0);
  }
  for (int i = 0; i < m; i++) {
    column_do(i, 0);
  }
  for (int i = 0; i < m; i++) {
    column_do(i, 1);
  }
  for (int i = 0; i < n; i++) {
    row_do(i, 1);
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (g[i][j] != mar[i][j][0]) {
        puts("-1");
        return;
      }
  int kind = ans[0] < ans[1] ? 0 : 1;
  printf("%d\n", ans[kind]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < rows[i][kind]; j++) printf("row %d\n", i + 1);
  for (int i = 0; i < m; i++)
    for (int j = 0; j < columns[i][kind]; j++) printf("col %d\n", i + 1);
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    init();
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        scanf("%d", &g[i][j]);
      }
    }
    solve();
  }
  return 0;
}